Question

The number of bacteria in a culture is given by the function N(t)=975e^.3twhere t is measured in hours.(a) What is the relative rate of growth of this bacterium population?(b) What is the initial population of the culture (at t=0)?(c) How many bacteria will the culture contain at time t=5?

Answer 1

EXPLANATION

Since we have the function:

$N(t)=975e^{0.3t}$

The standard form of a growth equation is the following:

$N(t)=N_0*e^{kt}$

Where N_0 = Initial population k= relative growth rate

a) The relative rate of growth is k=0.3

b) The initial population is 975 bacteria.

c) Plugging in the value t=5 into the expression:

$N(t)=975*e^{0.3*5}$

Multiplying numbers:

$N(t)=975*e^{1.5}$

Computing the exponent:

$N(t)=975*4.48$

Multiplying terms:

$N(t)=4368$

There will be 4368 bacteria at the time t=5